Question

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand. show the region in question, and interpret your result. $$\int_{0}^{4}(8-2 x) d x$$

Short answer

In this problem, we evaluated the definite integral $$\int_{0}^{4}(8-2 x) d x$$ by interpreting it as the area under the curve of the linear function $(8-2x)$ over the interval $[0, 4]$. We found that the region under the curve forms a triangular shape, with vertices at $(0, 8)$, $(4, 0)$, and $(0, 0)$. Using the formula for the area of a triangle, we calculated the area to be $16$ square units. This means that $$\int_{0}^{4}(8-2 x) d x = 16$$.

Step-by-step solution

Sketch the graph of the integrand function

To sketch the graph of the integrand function \((8-2x)\), we first notice it is a linear function with a slope of \(-2\) and a \(y\)-intercept of \(8\). The graph will be a straight line that starts from the point \((0,8)\) on the \(y\)-axis and moves downward at an angle.

Determine the geometric shape represented by the integrand

The graph of the integrand function is a straight line, and we need to find the area under it from \(x=0\) to \(x=4\). So, we will see where the line intersects the \(x\)-axis. To do this, set \(y=0\) and solve for \(x\): $$0=8-2x$$ $$2x=8$$ $$x=4$$ The line intersects the \(x\)-axis at \((4, 0)\). Therefore, the region under the graph from \(x=0\) to \(x=4\) forms a triangular shape, with vertex at points \((0, 8)\), \((4, 0)\), and \((0, 0)\).

Calculate the area of the triangular shape

To calculate the area of a triangle, we use the formula: $$area = \frac{1}{2} \cdot base \cdot height$$ For the triangle formed by our integrand, the base is \(4\) units long (from \(x=0\) to \(x=4\)) and the height is \(8\) units (since the \(y\)-intercept is at \((0, 8)\)). Now we can calculate the area: $$area = \frac{1}{2} \cdot 4 \cdot 8$$ $$area = 16$$

Interpret the result of the integral

In this context, the integral represents the area under the curve \((8-2x)\) between \(x=0\) and \(x=4\). We found out that this area is equal to \(16\) square units by using the geometric interpretation of the area. So, $$\int_{0}^{4}(8-2 x) d x = 16$$

Key concepts

Geometric Interpretation of Integrals

Understanding the geometric interpretation of integrals is fundamental when studying calculus. The process of integration can determine the accumulation of quantities, such as areas under curves. For instance, in the exercise \(\int_{0}^{4}(8-2x)dx\), the definite integral is calculated not by summing infinitesimal rectangles (as in Riemann sums), but by leveraging geometric shapes.

In this case, the linear function graphed creates a triangle when sketched, and the integral represents the total area under this line and above the x-axis, between x=0 and x=4. By seeing the integrand as a combination of simple shapes, like rectangles, triangles, or semicircles, we can calculate areas with basic geometry. This approach adds an intuitive layer to the abstract algebraic method, facilitating a deep understanding of what definite integrals represent.

Area Under the Curve

The area under the curve of a function is a central concept when interpreting definite integrals. When the exercise asks for the 'area under the curve', it refers to the space enclosed by the graph of the function, the x-axis, and the vertical lines corresponding to the limits of integration.

In our exercise, we visualize this as the area of a triangle formed by the graph of \(8-2x\), the x-axis, and the lines x=0 and x=4.

• Base of the triangle: the distance along the x-axis between the limits of integration (0 to 4).
• Height of the triangle: the value of the function at the y-intercept, where x=0.

Calculating the area of this triangle, \(\frac{1}{2} \cdot base \cdot height\), gives the value of the definite integral. This visual approach aids comprehension and provides a tangible way to appreciate what integration signifies - not just a number, but a measure of 'space' in the context of the graph.

Linear Function Graphing

Graphing a linear function is a straightforward process, pivotal in understanding problems involving definite integrals of such functions. A linear function is recognized by its constant rate of change, depicted as a straight line on a graph.

The integrand in our exercise, \(8-2x\), is a linear function characterized by a slope of -2 and a y-intercept of 8. The slope indicates the steepness and direction of the line, while the y-intercept is where the line crosses the y-axis.

• Plot the y-intercept (0, 8).
• Use the slope to determine another point, or when the line will cross the x-axis, in this case at (4, 0).
• Connect the points to extend the line in both directions.

Understanding how to sketch this effectively allows for the practical use of geometry in evaluating integrals and ensures that students can visualize the problem before tackling the algebra.